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How do find the derivative of #y=(1+cosx)/(1-cosx)#?

Answer 1

Emilia Aguilar

#y^' = (-2sinx)/(1-cosx)^2#

You can differentiate this function by using the quotient rule and the derivative of #cosx#, which is

#d/dx(cosx) = -sinx#

For a function that has the following formula:

#color(blue)(y = f(x)/(g(x)))#, where #g(x)!=0#

You can use the formula to find its derivative using the quotient rule.

#color(blue)(d/dx(y) = ([d/dx(f(x))] * g(x) - f(x) * d/dx(g(x)))/((g(x))^2)#

As for you,

#f(x) = 1 + cosx" "# and #" "g(x) = 1 - cosx#

Thus, you are able to write.

#d/dx(y) = ([d/dx(1 + cosx)] * (1-cosx) - (1 + cosx) * d/dx(1-cosx))/(1-cosx)^2#

#y^' = (-sinx * (1 - cosx) - (1 + cosx) * (-(-sinx)))/(1-cosx)^2#

#y^' = (-sinx + color(red)(cancel(color(black)(sinx * cosx))) - sinx - color(red)(cancel(color(black)(sinx * cosx))))/(1-cosx)^2#

#y^' = color(green)((-2sinx)/(1-cosx)^2)#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.